# Probability and Statistics (Test 2)

## Cat Entrance Exams : Mathematics Or Quantitative Aptitude

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Q.1
Direction: Study the given information carefully and answer the questions that follow:

A basket contains 5 red, 4 blue, 3 green stones.
If four balls are picked at random, what is the probability that two are blue and two are red?
A. 4/33
B. 3/35
C. 3/65
D. 5/33
E. None of these
Explaination / Solution:

Probability if two are Blue and two are Red
[(4C2*5C2)/12C4] = (6*10)/495 → 4/33

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Q.2

Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women

In how many ways this can be done if at least 5 men to have to be included in a committee

A. 251
B. 265
C. 167
D. 340
E. None of these
Explaination / Solution:

Number of Ways when if at least 5 men include in committee = 7C5*6C5+7C6*6C4+7C7*6C3

21*6+7*15+1*20 = 251

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Q.3
Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 5x2- 24x +16 = 0
II. 5y2 + 29y + 20 = 0
A. X>Y
B. X ≥Y
C. X
D. X≤Y
E. X = Y or the relationship cannot be established
Explaination / Solution:

I. 5x2- 24x +16 = 0
x = 4/5, 4
II. 5y2 + 29y + 20 = 0
y = (-4/5, -5)
So X>Y

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Q.4
Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women.
In how many ways of these committee the women are in majority
A. 45
B. 35
C. 110
D. 56
E. None of these
Explaination / Solution:

Number of ways when women are majority in committee = 6C6*7C4 = 1*35 = 35

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Q.5
Direction: Study the following information carefully and answer the questions that follow:

A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.
In how many ways this can be done if at least 2 trainees to have to be included in a committee
A. 340
B. 820
C. 560
D. 405
E. None of these
Explaination / Solution:

If at least 2 trainees include in committee
3C2*10C3 + 3C3*10C2 = 3*120+1*45
= 405

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Q.6
A coin is tossed three times, E : at most two tails , F : at least one tail. Find P(E|F)
A. 4/7
B. 3/7
C. 6/7
D. 2/7
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Q.7
If P(A) =, P(B) = and P(A ∪ B) = find P(A∩B)
A. 4/11
B. 4/13
C. 5/11
D. 5/17
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Q.8
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E)
A. P (E|F) =,P(F|E) =
B. P (E|F) =,P(F|E) =
C. P (E|F) =,P(F|E) =
D. P (E|F) =,P(F|E) =
Explaination / Solution:

We have ,
P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, then ,

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Q.9
A dice has 3 faces each bearing ‘ 2 ‘ and three faces each bearing ‘ 6 ‘. It is rolled once. The probability of showing up ‘a six ‘ is
A. 1
B. none of these
C. 1/6
D. 1/2
Explaination / Solution:

Total no. of outcomes = 6={2,2,2,6,6,6}

p(getting a six)=3/6=1/2

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Q.10
The probability that a card drawn at random from a pack of 52 cards is a king or a heart is
A. 1/ 52
B. 1/ 13
C. 16/ 52
D. 1/ 4